Android - how to find multiple views with common attribute

Question

I have a layout with multiple ImageViews, some of those images need to have the same onClickListener. I would like my code to be flexible and to be able to get a collection of those Views, iterate and add the listeners at run-time.

Is there a method like findViewById that will return a collection of Views rather than just a single one?

possible duplicate of [Find all views with tag?](http://stackoverflow.com/questions/5062264/find-all-views-with-tag) — johnkavanagh, Mar 05 '14 at 17:34

score 60 · Accepted Answer · edited Sep 29 '19 at 20:31

60

I've finally wrote this method (Updated thanks to @SuitUp (corrected username)):

private static ArrayList<View> getViewsByTag(ViewGroup root, String tag){
    ArrayList<View> views = new ArrayList<View>();
    final int childCount = root.getChildCount();
    for (int i = 0; i < childCount; i++) {
        final View child = root.getChildAt(i);
        if (child instanceof ViewGroup) {
            views.addAll(getViewsByTag((ViewGroup) child, tag));
        } 

        final Object tagObj = child.getTag();
        if (tagObj != null && tagObj.equals(tag)) {
            views.add(child);
        }

    }
    return views;
}

It will return all views that have android:tag="TAG_NAME" attribute. Enjoy ;)

edited Sep 29 '19 at 20:31

Morgoth

4,935
8
40
66

answered Jan 12 '12 at 07:38

Shlomi Schwartz

8,693
29
109
186

1

It does not collect ViewGroup Views. – SuitUp Jun 04 '12 at 20:13
4

You can use Objects.equal(tagObj, tag) for cleaner code and also allowing querying for views with no tags. Also you can make the method more general by letting the tag parameter be an object, since the tags may not be Strings. (And returning a List or Collection would be better practice). – Asaf Jan 23 '13 at 15:06
4

Pretty awesome, +1. I created a variation of this that returns the root view as well, if it has the tag. https://gist.github.com/orip/5566666 – orip May 13 '13 at 07:23
Why do you use final on each variable? See https://stackoverflow.com/questions/4279420/does-use-of-final-keyword-in-java-improve-the-performance – Lev Leontev Sep 29 '19 at 20:34

score 19 · Answer 2 · answered Jun 04 '12 at 20:13

Shlomi Schwartz's method has one flaw, it does not collect Views wchich are ViewGroups. Here is my fix:

private static ArrayList<View> getViewsByTag(ViewGroup root, String tag){
    ArrayList<View> views = new ArrayList<View>();
    final int childCount = root.getChildCount();
    for (int i = 0; i < childCount; i++) {
        final View child = root.getChildAt(i);
        if (child instanceof ViewGroup) {
            views.addAll(getViewsByTag((ViewGroup) child, tag));
        } 

        final Object tagObj = child.getTag();
        if (tagObj != null && tagObj.equals(tag)) {
            views.add(child);
        }

    }
    return views;
}

justadreamer · Answer 3 · 2012-01-11T14:43:42.627

6

There is no such method as findViewById that returns a group of views, but you can iterate over your views and set them an onClickListener like this:

for (int i = 0;i < layout.getChildCount();i++) {
  View child = layout.getChildAt(i);
  //you can check for view tag or something to see if it satisfies your criteria
  //child.setOnClickListener...
}

UPD: For recursive view hierarchy (this is example from a real project, where we do view refresh recursively, but instead you could just do whatever you want with the nested views):

private void recursiveViewRefresh(ViewGroup view) {
    for (int i = 0;i < view.getChildCount();i++) {
        View child = view.getChildAt(i);
        try {
            ViewGroup viewGroup = (ViewGroup) child;
            recursiveViewRefresh(viewGroup);
        } catch (ClassCastException e) {
            //just ignore the exception - it is used as a check
        }
        singleViewRefresh(child);
    }
}

edited Jan 11 '12 at 14:43

answered Jan 11 '12 at 10:08

justadreamer

2,410
2
21
24

I will go with that, although it is strange that there is no native method. – Shlomi Schwartz Jan 11 '12 at 12:20
What if the child is another layout? is it a recursion? – Shlomi Schwartz Jan 11 '12 at 13:47
yes you can do it recursively like this: (I will add it into the answer) – justadreamer Jan 11 '12 at 14:41
Thanks already implemented, I will post my code once Stack' will let me answer – Shlomi Schwartz Jan 11 '12 at 15:00
Check out the method I've added, I think it is a "good to have" util. Thanks again for your help ;) – Shlomi Schwartz Jan 12 '12 at 07:39

fernandohur · Answer 4 · 2017-01-07T11:55:55.220

This method provides a general way of obtaining views that match a given criteria. To use simply implement a ViewMatcher interface

private static List<View> getMatchingViews(ViewGroup root, ViewMatcher viewMatcher){
    List<View> views = new ArrayList<View>();
    final int childCount = root.getChildCount();
    for (int i = 0; i < childCount; i++) {
        final View child = root.getChildAt(i);
        if (child instanceof ViewGroup) {
            views.addAll(getMatchingViews((ViewGroup) child, viewMatcher));
        }

        if(viewMatcher.matches(child)){
            views.add(child);
        }
    }
    return views;
}

public interface ViewMatcher{
    public boolean matches(View v);
}

// Example1: matches views with the given tag
public class TagMatcher implements ViewMatcher {

    private String tag;

    public TagMatcher(String tag){
        this.tag = tag;
    }

    public boolean matches(View v){
        String tag = v.getTag();
        return this.tag.equals(tag);
    }

}


// Example2: matches views that have visibility GONE
public class GoneMatcher implements ViewMatcher {

    public boolean matches(View v){
        v.getVisibility() == View.GONE  
    }

}

TachikomaGT · Answer 5 · 2019-05-27T18:02:32.607

I will share my functional-style method with a filter, can be used with StreamSupport library.

@NonNull
public static <T> Function<View, Stream<T>> subViews(
    @NonNull final Function<View, Optional<T>> filter
) {
    return view -> RefStreams.concat(
        // If current view comply target condition adds it to the result (ViewGroup too)
        filter.apply(view).map(RefStreams::of).orElse(RefStreams.empty()),

        // Process children if current view is a ViewGroup
        ofNullable(view).filter(__ -> __ instanceof ViewGroup).map(__ -> (ViewGroup) __)
                .map(viewGroup -> IntStreams.range(0, viewGroup.getChildCount())
                        .mapToObj(viewGroup::getChildAt)
                        .flatMap(subViews(filter)))
                .orElse(RefStreams.empty()));
}

@NonNull
public static <T> Function<View, Optional<T>> hasType(@NonNull final Class<T> type) {
    return view -> Optional.ofNullable(view).filter(type::isInstance).map(type::cast);
}

@NonNull
public static Function<View, Optional<View>> hasTag(@NonNull final String tag) {
    return view -> Optional.ofNullable(view).filter(v -> Objects.equals(v.getTag(), tag));
}

Usage example:

Optional.ofNullable(navigationViewWithSubViews)
            .map(subViews(hasType(NavigationView.class)))
            .orElse(RefStreams.empty())
            .forEach(__ -> __.setNavigationItemSelectedListener(this));

score 1 · Answer 6 · answered Jul 31 '19 at 18:17

This is a modification of Shlomi Schwartz's answer above. All credit to them. I didn't like how the recursion looked, and needed to be able to regex match a string tag name.

  /**
   * Find all views with (string) tags matching the given pattern.
   *
   */
  protected static Collection<View> findViewsByTag(View root, String tagPattern) {
    List<View> views = new ArrayList<>();

    final Object tagObj = root.getTag();
    if (tagObj instanceof String) {
      String tagString = (String) tagObj;
      if (tagString.matches(tagPattern)) {
        views.add(root);
      }
    }

    if (root instanceof ViewGroup) {
      ViewGroup vg = (ViewGroup) root;
      for (int i = 0; i < vg.getChildCount(); i++) {
        views.addAll(findViewsByTag(vg.getChildAt(i), tagPattern));
      }
    }

    return views;
  }

For example:

Collection<View> itemNameViews = findViewsByTag(v, "^item_name_[0-9]+$");

score 0 · Answer 7 · answered Dec 27 '21 at 15:20

In kotlin we can do it is a very simplistic way as following (Kotlin Extension Function):

fun ViewGroup.findViewsByTag(tag: String): ArrayList<View> = ArrayList<View>().also {
    this.children.forEach { child ->
        if (child is ViewGroup) it.addAll(child.findViewsByTag(tag))
        if (child.tag == tag) it.add(child)
    }
}

Now, you can use it on any viewGroup. For example:

mViewGroup.findViewsByTag("my tag")

score 0 · Answer 8 · answered Jan 11 '12 at 10:07

0

@Override
protected void onClick(View view){
switch(view.getId()){

case R.id.whatEverImageViewId :
//....
break ;

case R.id.whatEverImageViewId 2 :
//....
break ;
....

and you can use for loop to add listeners

answered Jan 11 '12 at 10:07

Its not blank

3,055
22
37

score 0 · Answer 9 · edited Sep 16 '16 at 20:08

0

You can use switch() for multiple widgets.

switch(viewobject.getId()) {    
 case R.id.imageview1:    
   /* ... */    
   break;    
 case R.id.imageview2:    
   /* ... */      
   break;    
}

edited Sep 16 '16 at 20:08

Chris Stillwell

10,266
10
67
77

answered Jan 11 '12 at 13:07

Ranjit Mishra

440
4
8

Android - how to find multiple views with common attribute

9 Answers9

Linked